Question

Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is $7.3\times {10}^{–2}N{m}^{–1}$. Take the angle of contact to be zero and density of water to be $1.0\times {10}^{3}kg{m}^{–3}(g=9.8m{s}^{–2})$.

Answer 1

Diameter of the first bore, $d_1$ = 3.0 mm $=3\times {10}^{3}m$
Hence, the radius of the first bore $r_1=\frac{d_1}{2}=1.5\times {10}^{-3}m$
Diameter of the second bore, $d_2=6.0mm$
Hence, the radius of the second bore, $r_2=\frac{d_2}{2}=3\times {10}^{-3}m$
Surface tension of water, $s=7.3\times {10}^{–2}N{m}^{–1}$
Angle of contact between the bore surface and water, $\theta =0$
Density of water, $\rho =1.0\times {10}^{3}kg/m^{-3}$ Acceleration due to gravity, $g=9.8m/s^2$
Let $h_1$ and $h_2$ be the heights to which water rises in the first and second tubes respectively. These heights are given by the relations:
$h_1=\frac{2scos\theta }{r_1\rho g}.....\left(i\right)h_2=\frac{2scos\theta }{r_2\rho g}....\left(ii\right)$
The difference between the levels of water in the two limbs of the tube can be calculated as:
$=\frac{2scos\theta }{r_1\rho p}-\frac{2scos\theta }{r_2\rho g}=\frac{2scos\theta }{\rho g}[\frac{1}{r_1}-\frac{1}{r_2}]=\frac{2\times 7.3\times {10}^{-2}\times 1}{1\times {10}^3\times 9.8}[\frac{1}{1.5\times {10}^{-3}}-\frac{1}{3\times {10}^{-3}}]=4.966\times {10}^{-3}m=4.97mm$
Hence, the difference between levels of water in the two bores is 4.97 mm.